Properties

In this subsection, we will show the relation between Collatz function C(n) and its inverse operator, I(n). This helps us realize the relations among more natural numbers in the 3n + 1 backward iteration net.

Remark 2.3. For p, q∈ N, if C(p) = q, then p ∈ I(q).

proof : We will prove it by considering p.

Case 1: If p = 6k, for some k ∈ N, since C(p) = C(6k) = (6k)/2 = 3k = q, then I(q) = I(3k) = {2(3k)} = {6k}. Thus, p ∈ I(q).

Case 2: If p = 6k +1, for some k ∈ N∪{0}, since C(p) = C(6k+1) = 3(6k+1)+1 = 18k+4 = q, then I(q) = I(18k + 4) ={^(18k+4)₃ ⁻¹, 2(18k + 4)} = {6k + 1, 36k + 8}. Thus, p ∈ I(q).

Case 3: If p = 6k + 2, for some k ∈ N ∪ {0}, since C(p) = C(6k + 2) = (6k + 2)/2 = 3k + 1 = q, then I(q) = I(3k + 1) ={2(3k + 1)} = {6k + 2}. Thus, p ∈ I(q).

Case 4: If p = 6k+3, for some k ∈ N∪{0}, since C(p) = C(6k+3) = 3(6k+3)+1 = 18k+10 = q, then I(q) = I(18k +10) ={^(18k+10)₃ ⁻¹, 2(18k +10)} = {6k+3, 36k+20}. Thus, p ∈ I(q).

Case 5: If p = 6k + 4, for some k ∈ N ∪ {0}, since C(p) = C(6k + 4) = (6k + 4)/2 = 3k + 2 = q, then I(q) = I(3k + 2) ={2(3k + 2)} = {6k + 4}. Thus, p ∈ I(q).

Case 6: If p = 6k+5, for some k ∈ N∪{0}, since C(p) = C(6k+5) = 3(6k+5)+1 = 18k+16 = q, then I(q) = I(18k +16) ={^(18k+16)₃ ⁻¹, 2(18k +16)} = {6k+5, 36k+32}. Thus, p ∈ I(q).

Because Case 1 to Case 6 all hold, we claim the remark.

Remark 2.4. For p, q∈ N, if p ∈ I(q), then C(p) = q.

proof : We will prove it by considering q.

Case 1: If q = 6k, for some k ∈ N, since I(q) = I(6k) = {2(6k)} = {12k} and p ∈ I(q), then p = 12k. Thus, C(p) = C(12k) = (12k)/2 = 6k = q.

Case 2: If q = 6k + 1, for some k ∈ N ∪ {0}, since I(q) = I(6k + 1) = {2(6k + 1)} = {12k + 2}

and p∈ I(q), then p = 12k + 2. Thus, C(p) = C(12k + 2) = (12k + 2)/2 = 6k + 1 = q.

Case 3: If q = 6k + 2, for some k ∈ N, since I(q) = I(6k + 2) = {2(6k + 2)} = {12k + 4} and p∈ I(q), then p = 12k + 4. Thus, C(p) = C(12k + 4) = (12k + 4)/2 = 6k + 2 = q.

Case 4: If q = 6k + 3, for some k ∈ N, since I(q) = I(6k + 3) = {2(6k + 3)} = {12k + 6} and p∈ I(q), then p = 12k + 6. Thus, C(p) = C(12k + 6) = (12k + 6)/2 = 6k + 3 = q.

Case 5: If q = 6k + 4, for some k ∈ N, since I(q) = I(6k + 4) = {^(6k+4)₃ ⁻¹, 2(6k + 4)} = {2k + 1, 12k + 8} and p ∈ I(q), then p = 2k + 1 or p = 12k + 8. Now we consider the two conditions:

Subcase 1: If p = 2k + 1, then C(p) = C(2k + 1) = 3(2k + 1) + 1 = 6k + 4 = q.

Subcase 2: If p = 12k + 8, then C(p) = C(12k + 8) = (12k + 8)/4 = 6k + 4 = q.

By Subcase 1 and Subcase 2, the Case 5 holds.

Case 6: If q = 6k + 5, for some k ∈ N, since I(q) = I(6k + 5) = {2(6k + 5)} = {12k + 10} and p∈ I(q), then p = 12k + 10. Thus, C(p) = C(12k + 10) = (12k + 10)/2 = 6k + 5 = q.

Because Case 1 to Case 6 all hold, we claim the remark.

Proposition 2.5. If q ∈ N, then I(q) = {p| C(p) = q, p ∈ N}.

proof : By Remark 2.3 and Remark 2.4, we have: for p, q ∈ N, C(p) = q if and only if p∈ I(q). That is, I(q) = {p| C(p) = q}.

In Proposition 2.5, we know the relation between C(n) and I(n) under one transformation.

In the following, by use of Remark 2.3 and Remark 2.4, we will show the relation between C(n) and I(n) under k transformations.

Remark 2.6. For p, q, k ∈ N, if C^(k)(p) = q, then p∈ I^(k)(q).

proof : For p, q, k ∈ N, if C^(k)(p) = q, then there exists a finite trajectory P on C(n) with starting value p and length k, such that

P = (p, p₁, p₂, . . . , p_k−1, q), p_i = C⁽ⁱ⁾(p), i = 1, 2, 3, . . . , k− 1.

Since C(p) = p₁, by Remark 2.3, we have p ∈ I(p1). Similarly, we know p₁ ∈ I(p2), p₂ ∈ I(p3), . . ., pk−2 ∈ I(pk−1), and pk−1 ∈ I(q). By the relations, we have:

p∈ I(p1)⊆ I(I(p2))⊆ I(I(I(p3))) ⊆ · · · ⊆ I^(k−1)(pk−1)⊆ I^(k)(q).

Hence, p∈ I^(k)(q).

Remark 2.7. For p, q, k ∈ N, if p ∈ I^(k)(q), then C^(k)(p) = q.

proof : For p, q, k ∈ N, if p ∈ I^(k)(q), then there exists a path Q on I in β_k(q) with length k, such that

Q = (q → q1 → q2 → · · · → qk−1 → p), qi ∈ I⁽ⁱ⁾(q), i = 0, 1, 2, 3, . . . , k− 1.

Moreover, by Remark 2.4, we have q = C(q₁), q₁ = C(q₂), . . ., q_k−2 = C(q_k−1), and q_k−1 = C(p). Thus,

q = C(q₁) = C(C(q₂)) =· · · = C^(k−1)(q_k−1) = C^(k−1)(C(p)) = C^(k)(p).

Proposition 2.8. If q ∈ N, then I^(k)(q) ={p| C^(k)(p) = q, p∈ N}.

proof : By Remark 2.6 and Remark 2.7, we have: for p, q ∈ N, C^(k)(p) = q if and only if p∈ I^(k)(q). That is, I^(k)(q) ={p| C^(k)(p) = q}.

In Proposition 2.8, we realize the meaning of I^k(n), that is, by use of I^k(n), we can collect all the natural numbers which iterate to n under exactly k transformations on Collatz function.

In section 2, we introduce the 3n + 1 backward iteration net, and the relation of the natural numbers in the net. However, for our initial idea, we wonder if 1 can strength all the natural numbers set. Thus, in section 3, we focus on a particular 3n + 1 backward iteration net.

3 A Particular 3n + 1 Backward Iteration Net β

(8)

For our initial idea: if we let 1 be the starting value and make use of the 3n + 1 backward iteration formula I, can we strength all the natural numbers set N? Therefore, βk(1) is the main net what we interested in. However, we find that there is a trivial cycle 1→ 2 → 4 → 1 in β_k(1). For simplifying our work, the main net what we research transformed into β_k(8).

Why do we choose β_k(8)? Because observing β_k(1), we find that our initial idea is equal to the following idea: if we let 8 be the starting value and make use of the 3n + 1 backward iteration formula I, can we strength N \ {1, 2, 4}? Thus, in this section, βk(8) is the main 3n + 1 backward iteration net what we research. In this section, we will introduce β_k(8) and some properties in β_k(8).

3.1 Some Records of β

(8)

By the definition of 3n + 1 backward iteration net in section 2, β_k(8) is constructed by the starting value 8 and iterating k times on I, k ∈ N. In this section, we suppose k is a sufficient large natural number, then its partial framework as follows:

Figure 5: partial framework of β_k(8).

For convenience, we let I⁽ⁱ⁾(8) be L_i, called level i, i = 0, 1, 2,. . . , k. In the following, we list the first 12 levels of β_k(8):

L₀ : 8.

L₁ : 16.

L₂ : 5, 32.

L₃ : 10, 64.

L₄ : 3, 20, 21, 128.

L₅ : 6, 40, 42, 256.

L₆ : 12, 13, 80, 84, 85, 512.

L₇ : 24, 26, 160, 168, 170, 1024.

L₈ : 48, 52, 53, 320, 336, 340, 341, 2048.

L₉ : 96, 17, 104, 106, 640, 672, 113, 680, 682, 4096.

L₁₀: 192, 34, 208, 35, 212, 213, 1280, 1344, 226, 1360, 227, 1364, 1365, 8192.

L₁₁: 384, 11, 68, 69, 416, 70, 424, 426, 2560, 2688, 75, 452, 453, 2720, 454, 2728, 2730, 16384.

In the following, for our best, we record L₀ to L₅₉ in β_k(8), by Appendix A. Being ac-companied the increasing level, the total number L_i(N ) will be larger and larger. Therefore, we want to know the total number growing rate to realize β_k(8) more. We analysis the total number growing rate in β_k(8) as follows:

The total number growing rate L_i(N )/L_i−1(N ):

Owing to the proposition in 2.4: L_i+1(N )− Li(N ) = L_i(D) and L_i(D) ≥ 0. Therefore, we have L_i(N )/L_i−1(N ) ≥ 1. From L0 to L₁₀, the growing rate varies much, falls in [1, 2].

However, accompanied the increasing level, the growing rate seems more and more stable.

From L₂₅ to L₅₉, the growing rate all falls in (1.25, 1.27). Moreover, From L₄₃ to L₅₉, the growing rate all falls in (1.263, 1.265).

Although we just record L₀ to L₅₉ in β_k(8), by the analysis of the total number growing rate, we have some sense of total number in β_k(8). Moreover, according the growing rate trend, if k → ∞, then the total number in βk(8) has the chance to meet the N \ {1, 2, 4}.

3.2 Properties

In this subsection, by use of the remarks and propositions in 2.5, we will classify the properties in β_k(8) more precisely.

Remark 3.1. In β_k(8), 2ⁿ⁺³∈ Ln, for n = 0, 1, 2, . . . , k.

proof : We will prove it by mathematical induction.

(i) When n = 0, 2³ = 8∈ L0. True!

(ii) Suppose n = m, m < k, 2^m+3 ∈ Lm is true. Then by the formula I(n), we have:

∃m1 ∈ Lm+1 s.t. m1 = I(2^m+3) = 2· 2^m+3 = 2^m+4. Therefore, 2^m+4 ∈ Lm+1. True!

By mathematical induction, we prove it!

Remark 3.1 presents a trivial fact about 2-power natural numbers in β_k(8).

Remark 3.2. In β_k(8), the elements in L_i are distinct, i≤ k.

proof : If p∈ Li, then by Remark 2.7, we have: C⁽ⁱ⁾(p) = 8, i≤ k. Suppose there exist two finite trajectories X, Y with the same length i on Collatz function such that

X = (p, p₁, p₂, . . . , p_i−1, 8), C⁽ⁿ⁾(p) = p_n, n = 0, 1, 2, . . . , i− 1.

Y = (p, q₁, q₂, . . . , q_i−1, 8), C⁽ⁿ⁾(p) = q_n, n = 0, 1, 2, . . . , i− 1.

Since Collatz function, C⁽¹⁾(p) = p₁ = q₁, C⁽²⁾(p) = p₂ = q₂, similarly, p_i = q_i, for i = 1, 2, . . . , i− 1. That is, X ≡ Y . Thus, it shows that every element p in Li exists an unique finite trajectory with length i from p to 8 on Collatz function. Therefore, the elements in L_i are distinct.

Remark 3.3. In β_k(8), if p∈ Ls and p∈ Lt, then s = t.

proof : Suppose s̸= t and p ∈ Ls, p∈ Lt, without loss of generality, we let s < t ≤ k, then by Remark 2.7, we have: C^(s)(p) = 8 and C^(t)(p) = 8. Therefore, there exist a trajectory S with length s and a trajectory T with length t on Collatz function such that

S = (p, p₁, p₂, . . . , p_s−1, 8), C⁽ⁱ⁾(p) = p_i, i = 0, 1, 2, . . . , s− 1.

T = (p, q₁, q₂, . . . , q_t−1, 8), C⁽ⁱ⁾(p) = q_i, i = 0, 1, 2, . . . , t− 1.

Since Collatz function, C⁽¹⁾(p) = p₁ = q₁, C⁽²⁾(p) = p₂ = q₂, similarly, p_i = q_i, for i = 1, 2, . . . , s− 1. Thus, T can be written as

T = (p, p₁, p₂, . . . , p_s−1, 8, q_s+1, q_s+2, . . . , q_t−1, 8).

Now we consider the finite trajectory (8, q_s+1, q_s+2, . . . , q_t−1, 8) in T . Since C⁽¹⁾(8) = 4, C⁽²⁾(8) = 2, C⁽³⁾(8) = 1, C⁽⁴⁾(8) = 4, and so on. That is, q_i ∈ {1, 2, 4} for i > s. Thus, the finite trajectory (8, q_s+1, q_s+2, . . . , q_t−1, 8) does not exist. It is contradiction to our

assumption. Hence, we prove it!

Remark 3.4. In β_k(8), the elements in V (β_k(8)) are distinct.

proof: By Remark 3.3, we know that every element belongs to an unique level. Moreover, by Remark 3.2, we know the elements in a level are distinct. Therefore, the elements in

V (βk(8)) are distinct.

Theorem 3.5. For k ∈ N, there is no cycle in βk(8).

proof: By Remark 3.4, we know: the elements in V (β_k(8)) are distinct. Therefore, there is no cycle in β_k(8).

The Theorem 3.5 reveals a fact: For any n ∈ N \ {1, 2, 4}, if n can iterate to 1 on Collatz function, before getting to 1 at first time, then n don’t enter any cycle. It matches the fact we have known.

About our initial idea: if we let 8 be the starting value and make use of the 3n + 1 backward iteration formula, can we strength N \ {1, 2, 4}? Although we still don’t know whether the idea is right or not, we get some results in β_k(8). In 3.1, we know that if k → ∞, then the total number in β_k(8) have the chance to meet N \ {1, 2, 4}. In 3.2, we know all the natural numbers are different in βk(8). Moreover, in Theorem 3.5, we know there is no cycle in βk(8).

Through the research of β_k(8), we have another way to examine and interpret some facts about the 3n + 1 problem.

4 Simulation Backward Iteration Net

In this section, in order to know more about β_k(8), we construct a simulation expected matrix E to create a simulation backward net. We take the simulation backward net compared with β_k(8) on three parts: total numbers, total number growing rate, and the ratio of A to F in every level. We will introduce E and how we use E to simulate β_k(8). Moreover, we take another four different simulation expected value matrices to simulate β_k(8). In the end, we analysis the simulation results and make conclusions.

4.1 Simulation Expected Value Matrix E

In the subsection, we introduce the simulation expected value matrix E. The idea of E is from the relations of numbers in 3n + 1 backward iteration net. In section 2, we realize the natural number relations are a to b, b to d, c to f , e to d, f to f under one transformation on I. Based on the relations, we construct the simulation expected value matrix E. Especially, d may be to{a, b},{c, b}, or {e, b} under one transformation on I. Therefore, in E, we suppose the expected value of d to a, c, or e is 1/3 respectively under one transformation. It is more precise description below:

The first column is a-column. It shows that the expected value of “a iterating to b under one transformation” is 1. Moreover, the expected value of “a iterating to a, c, d, e, or f under one transformation on I” is 0 respectively.

(2) The second column:

[

0 0 0 1 0 0 ]T

.

The second column is b-column. It shows that the expected value of “b iterating to d under one transformation” is 1. Moreover, the expected value of “b iterating to a, b, c, e, or f under one transformation ” is 0 respectively.

(3) The third column:

[

0 0 0 0 0 1 ]T

.

The third column is c-column. It shows that the expected value of “c iterating to f under one transformation” is 1. Moreover, the expected value of “c iterating to a, b, c, d, or e under one transformation” is 0 respectively.

The fourth column is d-column. It shows that the expected value of “d iterating to b under one transformation” is 1. Moreover, the expected value of “d iterating to d, or f under one transformation” is 0. Especially, the expected value of “d iterating to a, c, or e under one transformation” is 1/3 respectively.

The fifth column is e-column. It shows that the expected value of “e iterating to d under one transformation” is 1. Moreover, the expected value of “e iterating to a, b, c, e, or f under one transformation” is 0 respectively.

The sixth column is f-column. It shows that the expected value of “f iterating to f under one transformation” is 1. Moreover, the expected value of “f iterating to a, b, c, d, or e, under one transformation” is 0 respectively.

X_a shows that we choose a as starting value and make the 3n + 1 backward iteration process. So do Xb, Xc, Xd, Xe, Xf. In the following, we take examples to explain how E and identity vectors work.

Example 4.1. E⁽¹⁾∗ Xd= [

1/3 1 1/3 0 1/3 0 ]T

.

It shows: If we choose d as starting value, under one transformation by E, the expected values of a is 1/3, of b is 1, of c is 1/3, of d is 0, of e is 1/3, of f is 0. That is, under one

It shows: If we choose d as starting value, under two transformations by E, the expected values of a is 0, of b is 1/3, of c is 0, of d is 4/3, of e is 0, of f is 1/3. That is, under two transformations by E, the next second level of starting value d consists 0· a, 1/3 · b, 0 · c, 4/3· d, 0 · e, 1/3 · f.

It shows: If we choose d as starting value, under three transformations by E, the expected values of a is 4/9, of b is 4/3, of c is 4/9, of d is 1/3, of e is 4/9, of f is 1/3. That is, under three transformations by E, the next third level of starting value d consists 4/9· a, 4/3 · b, 4/9· c, 1/3 · d, 4/9 · e, 1/3 · f.

In the following, we will introduce that how we use E to simulate β_k(8).